**AND xor OR**

### Problem Statement :

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value of maximum possible value of Si.

### Solution :

` ````
Solution in C :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
int main() {
int n;
cin>>n;
int * arr;
arr = new int[n];
for(int i=0;i<n;i++)cin>>arr[i];
stack<int> S;
int m = 0;
for(int i=0;i<n;i++)
{
while(true)
{
if(S.empty())
{
S.push(arr[i]);
break;
}
else if(S.top() < arr[i] )
{
m = max(m, S.top()^arr[i]);
S.push(arr[i]);
break;
}
else
{
S.pop();
}
}
}
stack<int> S2;
for(int i=n-1;i>=0;i--)
{
while(true)
{
if(S.empty())
{
S.push(arr[i]);
break;
}
else if(S.top() < arr[i] )
{
m = max(m, S.top()^arr[i]);
S.push(arr[i]);
break;
}
else
{
S.pop();
}
}
}
cout<<m<<endl;
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++){
arr[i] = in.nextInt();
}
Stack<Integer> stack = new Stack<Integer>();
int result = 0;
for(int i = 0; i < n; i++){
while(true){
if(stack.isEmpty()){
stack.add(arr[i]);
break;
} else if(stack.peek() < arr[i]){
result = Math.max(result, stack.peek() ^ arr[i]);
stack.add(arr[i]);
break;
} else{
stack.pop();
}
}
}
stack = new Stack<Integer>();
for(int i = n - 1; i >= 0; i--){
while(true){
if(stack.isEmpty()){
stack.add(arr[i]);
break;
} else if(stack.peek() < arr[i]){
result = Math.max(result, stack.peek() ^ arr[i]);
stack.add(arr[i]);
break;
} else{
stack.pop();
}
}
}
System.out.println(result);
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
long long int sol(int A, int B)
{
return (((A & B) ^ (A | B)) & (A ^ B));
}
long long int inter(long long int arr[], int n)
{
int i,j;
long long int min1 = 9999999999, min2 = 9999999999, max = -1;
for(i=0;i<n-1;i++)
{
if(sol(arr[i], arr[i+1]) > max)
max = sol(arr[i], arr[i+1]);
if(i+2 <= n-1)
if(arr[i+1] > arr[i] && arr[i+1] > arr[i+2] && sol(arr[i], arr[i+2]) > max)
max = sol(arr[i], arr[i+2]);
}
return max;
}
int main()
{
int n,i;
scanf("%d", &n);
long long int arr[n];
for(i=0;i<n;i++)
{
scanf("%lli", &arr[i]);
}
printf("%lli", inter(arr, n));
return 0;
}
In Python3 :
n_elements = int(input())
data = list(map(int, input().split()))
assert len(data) == n_elements
res = 0
minima = [data[0]]
for i in range(1, n_elements):
d = data[i]
while len(minima):
x = minima.pop()
res = max(d ^ x, res)
if x < d:
minima.append(x)
break
minima.append(d)
print(res)
```

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